Ncert Solutions Chemistry Class 11th
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11.14. Suggest reasons why the B-F bond lengths in BF3 (130 pm) and BF4– (143 pm) differ.
11.14. Suggest reasons why the B-F bond lengths in BF3 (130 pm) and BF4– (143 pm) differ.
The difference in bond length is due to the difference in the state of hybridisation. In BF3 'B' is sp2 hybridised and in BF4– 'B' is sp3 hybridised.
Upvote 
1.5. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.
1.5. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.
1.5
Since the molar mass of sodium acetate is 82.0245 g mol-1.
⇒ 1 mol of CH3COONa has a mass = 82.0245 g
∴Mass of sodium acetate (CH3COONa) required to make 500 ml of 0.375 molar aqueous solution
= 0.375 x 0.5 mol x 82.0245 g mol-1
= 15.3795 g = 15.380 g
Upvote 11.13.
- Rationalise the given statements and give chemical reactions:
- Lead (II) chloride reacts with Cl2 to give PbCl4.
- Lead (IV) chloride is highly unstable towards heat.
- Lead is known not to form an iodide PbI4.
11.13.
- Rationalise the given statements and give chemical reactions:
- Lead (II) chloride reacts with Cl2 to give PbCl4.
- Lead (IV) chloride is highly unstable towards heat.
- Lead is known not to form an iodide PbI4.
- PbCl2 + Cl2→ PbCl4.
This is because Pb can show +2 oxidation state more easily than +4 due to inert pair effect. - PbCl4→ PbCl2 + Cl2
Because Pb2+ is more stable than Pb4+ due to inert pair effect. - PbI4does not exist because I- ion being a powerful reducing agent reduces Pb4+ ion to Pb
Upvote 11.12. Explain the difference in properties of diamond and graphite on the basis of their structures.
11.12. Explain the difference in properties of diamond and graphite on the basis of their structures.
Diamond has a crystalline lattice where each carbon atom undergoes sp3 hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. The C–C bond length is 154 pm. The structure extends in space and produces a rigid three- dimensional network of carbo
Upvote 1.4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
1.4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
1.4 In order to answer the question, we need to know the balanced equation for the combustion of carbon in dioxygen/air, which can be written as:
C (s) + O2 (g) à CO2 (g)
(i) We can see from the above equation,
1 mole of carbon reacts with 1 mole of oxygen to produce 1 mol of carbon dioxide.
In air, c
Upvote 11.11. What is the state of hybridisation of carbon in:
(a) CO32- (b) Diamond (c) Graphite?
11.11. What is the state of hybridisation of carbon in:
(a) CO32- (b) Diamond (c) Graphite?
(a) CO32- (sp2) (b) Diamond (sp3) (c) Graphite (sp2)
Upvote 1.3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1 % dioxygen by mass.
1.3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1 % dioxygen by mass.
1.3 We are given that the percentage of iron by mass is 69.9% and the percentage of oxygen by mass is 30.1%.
Since we have relative moles of both elements, we can calculate the simpler molar ratio of iron to oxygen
= 1.25: 1.88
Divide by the smaller value to both
= 1.25/1.25: 1.88/1.25
= 1: 1.5
= 2: 3
So,
Upvote 11.9. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.
11.9. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.
Electron deficient compounds are those in which the central atom in their molecule has the tendency to accept one or more electron pairs. They are also known as Lewis acid. BCl3 and SiCl4 both are electron deficient species.
Since, in BCl3, B atom has only six electrons. Therefore, it is an
Upvote 11.8. Write reactions to justify amphoteric nature of aluminium.
11.8. Write reactions to justify amphoteric nature of aluminium.
Aluminium reacts with acid as well as base. This shows amphoteric nature of aluminium.
2Al (s) + 6HCl (dil.) →2AlCl3 (aq) + 3H2 (g)
2Al (s) + 2NaOH (aq) + 6H2O (l) →2Na+ [Al (OH)4]– (aq) + 3H2 (g)
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